Hence, $0.01$ mole of $NaOH$ will be completely neutralised by $0.01$ mole of $HCL$.
$\Rightarrow\; NaOH$ left unneutralised $=0.1-0.01=0.09$ mol As equal volumes of two solutions are mixed,
$[\mathrm{OH}]^{-}=\frac{0.09}{2}=0.045 \mathrm{M} $
$ \Rightarrow \mathrm{pOH}=-\log (0.045)=1.35 $
$\therefore \mathrm{pH}=14-1.35=12.65$